# Question #47c93

##### 1 Answer

#### Answer:

Here's what I got.

#### Explanation:

Your tool of choice for this problem will be the **Arrhenius equation**, which establishes a relationship between the **rate constant** of a reaction,

#color(blue)(k = A * "exp"(-E_a/(RT))" " " "color(purple)((1))" "# , where

*pre-exponential factor*

**activation energy** of the reaction

*universal gas constant*, in this context given as

Now, the *Arrhenius parameters* are

Since you're basically dealing with one equation and two unknowns, you will have to use the information given to find a way to eliminate one of the two Arrhenius parameters first.

You can do that by writing the Arrhenius equation in non-exponential form

#ln(k) = ln[A * "exp"(-E_a/(RT))]#

This is equivalent to

#ln(k) = ln(A) + ln["exp"(-E_a/(RT))]#

#ln(k) = ln(A) - E_a/(RT)#

Now, notice that the problem provides you with information about the value of the rate constant *at two different temperatures*

#T_1 = 273.15 + 24 = "297.15 K"#

#T_2 = 273.15 + 37 = "310.15 K"#

This means that you can write

#ln(k_1) = ln(A) - E_a/(R * T_1) -># for#k_1# and#T_1#

#ln(k_2) = ln(A) - E_a/(R * T_2) -># for#k_2# and#T_2#

Notice that you can get rid of the

#ln(k_1) - ln(k_2) = color(red)(cancel(color(black)(ln(A)))) - color(red)(cancel(color(black)(ln(A)))) - E_a/R * (1/T_1 - 1/T_2)#

This will get you

#color(blue)(ln(k_1/k_2) = - E_a/R * (1/T_1 - 1/T_2))#

You can now find the activation energy of the reaction - keep in mind that activation energy **must** carry a *positive sign*, since it expresses energy *required*

#E_a = - (R * ln(k_1/k_2))/(1/T_1 - 1/T_2)#

#E_a = - (8.3145 "J"/("mol" * color(red)(cancel(color(black)("K")))) * ln( (1.70 * 10^(-2) color(red)(cancel(color(black)("L mol"^(-1)"s"^(-1)))))/(2.01 * 10^(-2) color(red)(cancel(color(black)("L mol"^(-1)"s"^(-1)))))))/((1/297.15 - 1/310.15) 1/color(red)(cancel(color(black)("K"))))#

#E_a = "9873.5 J/mol"#

To get the value of the pre-exponential factor, pick a value for

For

#k_1 = A * "exp"(-E_a/(RT_1))#

This will get you

#A = k_1/("exp"(-E_a/(RT)))#

Notice that since the exponential has to be **unitless**,

#A = (1.70 * 10^(-2)"L mol"^(-1)"s"^(-1))/("exp"(- (9873.5 color(red)(cancel(color(black)("J")))/color(red)(cancel(color(black)("mol"))))/(8.3145color(red)(cancel(color(black)("J")))/(color(red)(cancel(color(black)("mol"))) * color(red)(cancel(color(black)("K")))) * 297.15color(red)(cancel(color(black)("K"))))))#

#A = "0.92475 L mol"^(-1)"s"^(-1)#

I'll leave the answers rounded to three sig figs.

#E_a = color(green)("9.87 kJ/mol") -># expressed in kilojoules per mole

#A = color(green)("0.925 L mol"^(-1)"s"^(-1))#